Is (tan(X)-sin(X))/(sin^3(X))=(sec(X))/(1+cos(X)) ?

The answer to whether (tan(X)-sin(X))/(sin^3(X))=(sec(X))/(1+cos(X)) is True

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Popular >

prove (tan(X)-sin(X))/(sin^3(X))=(sec(X))/(1+cos(X))

Solution

prove

\frac{tan ( X ) - sin ( X )}{sin ^{3} ( X )} = \frac{sec ( X )}{1 + cos ( X )}

Solution

True

Solution steps

\frac{tan ( X ) - sin ( X )}{sin ^{3} ( X )} = \frac{sec ( X )}{1 + cos ( X )}

Manipulating left side

\frac{tan ( X ) - sin ( X )}{sin ^{3} ( X )}

Express with sin, cos

\frac{- sin ( X ) + tan ( X )}{sin ^{3} ( X )}

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= \frac{- sin ( X ) + \frac{s i n ( X )}{c o s ( X )}}{sin ^{3} ( X )}

Simplify

\frac{- sin ( X ) + \frac{s i n ( X )}{c o s ( X )}}{sin ^{3} ( X )} : \frac{- cos ( X ) + 1}{sin ^{2} ( X ) cos ( X )}

\frac{- sin ( X ) + \frac{s i n ( X )}{c o s ( X )}}{sin ^{3} ( X )}

Join

- sin (X) + \frac{sin ( X )}{cos ( X )} : \frac{- sin ( X ) cos ( X ) + sin ( X )}{cos ( X )}

- sin (X) + \frac{sin ( X )}{cos ( X )}

Convert element to fraction:

sin (X) = \frac{sin ( X ) cos ( X )}{cos ( X )}

= - \frac{sin ( X ) cos ( X )}{cos ( X )} + \frac{sin ( X )}{cos ( X )}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- sin ( X ) cos ( X ) + sin ( X )}{cos ( X )}

= \frac{\frac{- s i n ( X ) c o s ( X ) + s i n ( X )}{c o s ( X )}}{sin ^{3} ( X )}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{- sin ( X ) cos ( X ) + sin ( X )}{cos ( X ) sin ^{3} ( X )}

Factor out common term

sin (X)

= \frac{sin ( X ) ( - cos ( X ) + 1 )}{cos ( X ) sin ^{3} ( X )}

Cancel the common factor:

sin (X)

= \frac{- cos ( X ) + 1}{sin ^{2} ( X ) cos ( X )}

= \frac{- cos ( X ) + 1}{sin ^{2} ( X ) cos ( X )}

= \frac{1 - cos ( X )}{cos ( X ) sin ^{2} ( X )}

Rewrite using trig identities

\frac{1 - cos ( X )}{cos ( X ) sin ^{2} ( X )}

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= \frac{1 - cos ( X )}{cos ( X ) ( 1 - cos ^{2} ( X ) )}

Simplify

\frac{1 - cos ( X )}{cos ( X ) ( 1 - cos ^{2} ( X ) )} : \frac{1}{cos ( X ) ( cos ( X ) + 1 )}

\frac{1 - cos ( X )}{cos ( X ) ( 1 - cos ^{2} ( X ) )}

Factor

1 - cos^{2} (X) : - (cos (X) + 1) (cos (X) - 1)

1 - cos^{2} (X)

Factor out common term

- 1

= - (cos^{2} (X) - 1)

Factor

cos^{2} (X) - 1 : (cos (X) + 1) (cos (X) - 1)

cos^{2} (X) - 1

Rewrite

1

1^{2}

= cos^{2} (X) - 1^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

cos^{2} (X) - 1^{2} = (cos (X) + 1) (cos (X) - 1)

= (cos (X) + 1) (cos (X) - 1)

= - (cos (X) + 1) (cos (X) - 1)

= - \frac{1 - cos ( X )}{cos ( X ) ( cos ( X ) + 1 ) ( cos ( X ) - 1 )}

Cancel

- \frac{1 - cos ( X )}{cos ( X ) ( cos ( X ) + 1 ) ( cos ( X ) - 1 )} : \frac{1}{cos ( X ) ( cos ( X ) + 1 )}

- \frac{1 - cos ( X )}{cos ( X ) ( cos ( X ) + 1 ) ( cos ( X ) - 1 )}

cos (X) - 1 = - (- cos (X) + 1)

= - \frac{- cos ( X ) + 1}{- cos ( X ) ( cos ( X ) + 1 ) ( - cos ( X ) + 1 )}

Refine

= \frac{1 - cos ( X )}{cos ( X ) ( cos ( X ) + 1 ) ( - cos ( X ) + 1 )}

Cancel the common factor:

1 - cos (X)

= \frac{1}{cos ( X ) ( cos ( X ) + 1 )}

= \frac{1}{cos ( X ) ( cos ( X ) + 1 )}

= \frac{1}{cos ( X ) ( cos ( X ) + 1 )}

= \frac{1}{cos ( X ) ( cos ( X ) + 1 )}

Rewrite using trig identities

Use the basic trigonometric identity:

\frac{1}{cos ( x )} = sec (x)

\frac{sec ( X )}{( cos ( X ) + 1 )}

Simplify

\frac{sec ( X )}{( cos ( X ) + 1 )} : \frac{sec ( X )}{cos ( X ) + 1}

\frac{sec ( X )}{( cos ( X ) + 1 )}

Remove parentheses:

(a) = a

= \frac{sec ( X )}{cos ( X ) + 1}

\frac{sec ( X )}{cos ( X ) + 1}

\frac{sec ( X )}{cos ( X ) + 1}

= \frac{sec ( X )}{1 + cos ( X )}

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is (tan(X)-sin(X))/(sin^3(X))=(sec(X))/(1+cos(X)) ?
The answer to whether (tan(X)-sin(X))/(sin^3(X))=(sec(X))/(1+cos(X)) is True